And what possible reason could there be for spending money on an ASIC at this early stage? It's only ultrasonic - nothing that couldn't be handled by an FPGA/DSP and off-the-shelf analogue parts. $20M buys a lot of those.

No, it's so they can have some proprietary bullshit integrated into it like every garbage hippie IOT product. It must force you into some overpriced cloud service that steals all of your data and spies on you dressing.

And what possible reason could there be for spending money on an ASIC at this early stage? It's only ultrasonic - nothing that couldn't be handled by an FPGA/DSP and off-the-shelf analogue parts. $20M buys a lot of those.

Is the ASIC to do the signal processing, or is it a power device to drive the huge number of transmitter cells they have (or even combine energy from the array of receivers at the phone)?

And what possible reason could there be for spending money on an ASIC at this early stage? It's only ultrasonic - nothing that couldn't be handled by an FPGA/DSP and off-the-shelf analogue parts. $20M buys a lot of those.

Is the ASIC to do the signal processing, or is it a power device to drive the huge number of transmitter cells they have (or even combine energy from the array of receivers at the phone)?

As far as I understood, the ASIC design houses the transducers. Note, they are transducers/transceiver, with a possible intent of data transfer. I guess Nyquist was also a linear thinker...

The transmitter sends out a collimated beam of 1KW/M^2, e.g. 0.1W per cm^2 of optical power.There's some range, which means about 50% loss (Air is lossy), and conversion of ultrasound to electricity can be done at some 10% efficiency (commercial 1-2%) Recievers have to work at some angles, I believe 45 degrees is minimal, so you need to reduce this by 30% more.So that's about 3mW per cm^2 receiver, or 20mW/Inch^2

A mouse consumes about 1mW of power to run it and process the data, so the receiver can be small enough.

Now do the same for a phone.... A phone uses 5W to charge.... so you will need some 250 Inch^2 to charge it, you will need a 20" screen phablets (or phaTVs?) will easily accomodate it.

The transmitter sends out a collimated beam of 1KW/M^2, e.g. 0.1W per cm^2 of optical power.There's some range, which means about 50% loss (Air is lossy), and conversion of ultrasound to electricity can be done at some 10% efficiency (commercial 1-2%) Recievers have to work at some angles, I believe 45 degrees is minimal, so you need to reduce this by 30% more.So that's about 3mW per cm^2 receiver, or 20mW/Inch^2

A mouse consumes about 1mW of power to run it and process the data, so the receiver can be small enough.

Now do the same for a phone.... A phone uses 5W to charge.... so you will need some 250 Inch^2 to charge it, you will need a 20" screen phablets (or phaTVs?) will easily accomodate it.

Not sure how you get that number for the phone. If at 150 dB that needs 0.005m^2 to get 5W, let's be super generous and say 10W with 50% efficiency of receive conversion, so 0.01m. An iPhone X is about 7.5 by 15cm so 0.01125m^2.

Wow, isn't that lucky. 150 dB works out perfectly, under perfect conditions and a receive efficiency similar to what you see in one of those pitch graphs, to charge an iPhone X rapidly. So if conditions aren't perfect, or 150dB is deemed unsafe, or receive efficiency can't approach 50%, then it's downhill from there.

Oh, and that's from a 0.33m^2 transmitter, say at 50% efficiency, so 666W used, so <1% efficiency.

The transmitter sends out a collimated beam of 1KW/M^2, e.g. 0.1W per cm^2 of optical power.There's some range, which means about 50% loss (Air is lossy), and conversion of ultrasound to electricity can be done at some 10% efficiency (commercial 1-2%) Recievers have to work at some angles, I believe 45 degrees is minimal, so you need to reduce this by 30% more.So that's about 3mW per cm^2 receiver, or 20mW/Inch^2

A mouse consumes about 1mW of power to run it and process the data, so the receiver can be small enough.

Now do the same for a phone.... A phone uses 5W to charge.... so you will need some 250 Inch^2 to charge it, you will need a 20" screen phablets (or phaTVs?) will easily accomodate it.

Not sure how you get that number for the phone. If at 150 dB that needs 0.005m^2 to get 5W, let's be super generous and say 10W with 50% efficiency of receive conversion, so 0.01m. An iPhone X is about 7.5 by 15cm so 0.01125m^2.

Wow, isn't that lucky. 150 dB works out perfectly, under perfect conditions and a receive efficiency similar to what you see in one of those pitch graphs, to charge an iPhone X rapidly. So if conditions aren't perfect, or 150dB is deemed unsafe, or receive efficiency can't approach 50%, then it's downhill from there.

Oh, and that's from a 0.33m^2 transmitter, say at 50% efficiency, so 666W used, so <1% efficiency.

Unlike you, I'm not an ultrasound expert, so.... But I'm a practical person.... and believe products should be designed to be working, not to be lab experiments published in a research paper.

For a phone to charge, you need 5W of power, in normal conditions, that are actually competitive to available solutions, e.g. better compared to a Qi pad or usb cable.

The way I interpret the above is working when tilted to some degree (45 degrees sounds reasonable, although I would prefer 60-70)At some distance (say 5-10 feet)Even at 100% humidity, and cold/hot weather.A phone that's as clean as a normal phone is, possibly with some of the area obscured.It has to be safe, legal, and not annoying (to me, my kids or my pets, also, for it to be in my livingroom, it should not make a fan noise)

the number 1KW/M^2 is taken from uBeam's ppt, it's comparable to 150dBI know it's possible to focus that power to a smaller spot with a phased array, but considering uBeam must be transmitting at a safe level (I'll take their word 150dB is safe and legal, althogh I know it to be not true), the beam cannot exceed this level at any point, in any weather conditions. So, if attenuation for dry air for 10-20dB higher compared to moist air, then the power density at the receiver can't exceed ~140dB acoustic.

Taking into account your 50% conversion efficiency (I believe you, although I never seen anything close) that's equivalent to 5mW/cm^2

The effective area of iPhone X tilted at 45 degrees is 70 cm^2 - that's not enough (if it were enough a solar cell would be enough to keep the phone charged, no transmitter needed).

It has to be safe, legal, and not annoying (to me, my kids or my pets, also, for it to be in my livingroom, it should not make a fan noise)

the number 1KW/M^2 is taken from uBeam's ppt, it's comparable to 150dB

Don't worry. After charging you phone at 150dB you DEFINITELY won't hear any fan noise

Please remember, Paul is doing generous calculations with reasonable/optimistic assumptions to show that this is impractical and extremely inefficient as proposed. Arguing over small changes in assumptions doesn't change the essential message!

I love the picture of the uBeam soundbar along the bottom of the monitor.I can't wait to be able to work all day directly in front of a 150dBm ultrasonic transmitter.

Don't worry, the uBean sound bar will only transmit the US power in narrow beams to devices actively requesting power using the uBean proprietary white rectangle power request protocol, - so everything will be fine.

Not sure how you get that number for the phone. If at 150 dB that needs 0.005m^2 to get 5W, let's be super generous and say 10W with 50% efficiency of receive conversion, so 0.01m. An iPhone X is about 7.5 by 15cm so 0.01125m^2.

Unlike you, I'm not an ultrasound expert, so.... But I'm a practical person.... and believe products should be designed to be working, not to be lab experiments published in a research paper.

Thanks for explaining the methodology. It's mostly correct but what I suggest you look at is not the system from the Transmitter to the Receiver, but vice versa - the receiver is the fixed size, instead the question is "what's the size of transmitter needed to provide this power?". uBeam have stated the transmitter is 60x60cm for their "up to 8 Watts at 1 meter" chart, and somewhere between a small amount and all of that area can be used to target a phone. Even though there is loss between the transmitter and receiver, more transmit area can be applied to compensate for that loss and maintain the same power incident on the receiver, up until the point you run out of transmit area. Power delivered remains constant, but you lose efficiency - and that's the number that's hidden unless you go digging.

So I think you've run your numbers backwards. You've more done a "how much transmit area do I need to power a phone sized area at 5W?". You've come up with 1400 cm^2. The uBeam full panel is 3600 m^2. Ballpark similar, especially if you think they limit drive to 145dB (3x less power)

Once again we come back to the same point, which is - of course you can transmit power ultrasonically, but what about safety, efficiency, cost, and practicality?

And, as someone else points out, I tend to give calculations here that give best case numbers to a) keep it simple and b) show even in best case it's not great. Believe me, I know the difference between research and a practical product.

I waiting for the announcement of uBeam transmitters on power poles beside the road for charging cars.Can someone please troll tweet Meredith with that idea to see if it's practical? Makes sense after all, the power infrastructure is already there, and their beamforming tech has got to be great for targeting passing cars!

I waiting for the announcement of uBeam transmitters on power poles beside the road for charging cars.Can someone please troll tweet Meredith with that idea to see if it's practical? Makes sense after all, the power infrastructure is already there, and their beamforming tech has got to be great for targeting passing cars!

Someone drives by in a Tesla truck and the pole transformer just explodes.

Not sure how you get that number for the phone. If at 150 dB that needs 0.005m^2 to get 5W, let's be super generous and say 10W with 50% efficiency of receive conversion, so 0.01m. An iPhone X is about 7.5 by 15cm so 0.01125m^2.

... more transmit area can be applied to compensate for that loss and maintain the same power incident on the receiver, up until the point you run out of transmit area.

I'm not sure this is true,Assuming some level power, say 145dB, 150dB, whatever.... is deemed unsafe The power level at the receiver side cannot exceed this safe limit, can it?

So If the effective phone size is 70cm^2, and the maximal power density that's still safe (on either side) is 150dB than 7W can really be transmitted.

But...

Assume the transmitter is actually transmitting a few watts in 0% humidity.A day later, a receiver receives 1% of the power, can the transmitter increase it's output 100X?

It can't, for two reasons1. To do that it must know what is the exact reason for the decline in received power is, if the reason is that my body absorbs 99% of the power I hope it's not going to increase it. Knowing the exact reason why something delivers less power, takes a human a day's work in the lab and is beyond the current capabilities of devices.2. Such dynamic range means your costs are X100 higher, if your "dry day" cost is anywhere above $10, your wet day costs are sky high.